Optimal fee structures in hedge funds

Abstract

This paper proposes a framework to analyze hedge funds fee arrangements in which the portfolio construction is determined by the hedge fund manager and the fees are determined via an optimal equilibrium between the manager and the investor. In this setting, fees include management fees ($\alpha$) and performance fees ($\beta$). We select the paradigm of Expected Utility Theory to determine the managers optimal strategy. Benefiting from the dependence of the optimal terminal payoff on the fee structure, we explore two criteria producing an equilibrium fee mutually satisfying the investor and the manager, one based on Pareto Optimality and the second on negotiable regions. The former also leads to a Pareto efficient frontier of fee structures. We obtain evidence that the popular fee structure of $(\alpha , \beta )=(2\%, 20\%)$ is not an equilibrium fee between manager and investor. Although such equilibrium heavily depends on risk aversion levels and market conditions, the pair $(\alpha , \beta )=(0.5\%, 30.7\%)$ stands out as a fair choice. Moreover, the expected utility of the investor is not monotone in the performance fee for many market conditions. In other words, we prove that fee arrangements which include performance fees are usually beneficial for the investor.

Appendices

Appendix 1: Figures

See Figs. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24 and 25.

Appendix 2: Mathematical proofs

Proof of Lemma1:

We show that (13) is equivalent to (15). Therefore, we use the following result.

$$\begin{aligned} {\mathbb {E}}\left[ {\tilde{Z}}_T^{k}{1}_{\{a<{\tilde{Z}}_T<b\}}|{ \mathcal {F}}_t\right]&={\tilde{Z}}_t^{k}{\mathbb {E}}\left[ \left( \frac{{ \tilde{Z}}_T}{{\tilde{Z}}_t}\right) ^{k}{1}_{\{\frac{a}{{\tilde{Z}}_t}< \frac{{\tilde{Z}}_T}{{\tilde{Z}}_t}<\frac{b}{{\tilde{Z}}_t}\}}\right] = {\tilde{Z}}_t^{k}{\mathbb {E}}\left[ \left( \frac{{\tilde{Z}}_T}{{ \tilde{Z}}_t}\right) ^{k}{1}_{\{d_2<\frac{W_{T-t}}{\sqrt{T-t}}<d_1\}}\right] \end{aligned}$$

with

$$\begin{aligned} d_1=-\frac{{\hbox {ln}}(\frac{a}{{\tilde{Z}}_t})+\left( r+\frac{\gamma ^2}{2}\right) (T-t)}{ \gamma \sqrt{T-t}} \text{ and } d_2=-\frac{{\hbox {ln}}(\frac{b}{{\tilde{Z}}_t})+\left( r+\frac{\gamma ^2}{2} \right) (T-t)}{\gamma \sqrt{T-t}} \end{aligned}$$

Hence,

$$\begin{aligned} {\mathbb {E}}\left[ {\tilde{Z}}_T^{k}{1}_{\{a<{\tilde{Z}}_T<b\}}|{\mathcal {F}}_t\right]&= { \tilde{Z}}_t^{k} \int _{d_2}^{d_1}{\mathrm{e}}^{-k\gamma \sqrt{T-t}\cdot x-k(r+ \frac{\gamma ^2}{2})(T-t)}\frac{1}{\sqrt{2 \pi }}{\mathrm{e}}^{-\frac{1}{2}x^2}\mathrm {d}x \nonumber \\&= {\tilde{Z}}_t^k {\mathrm{e}}^{-k(r+\frac{\gamma ^2}{2})(T-t)}\int _{d_2}^{d_1} \frac{1}{\sqrt{2\pi }}{\mathrm{e}}^{-\frac{1}{2}(x+k\gamma \sqrt{T-t})^2+\frac{k^2}{2} \gamma ^2(T-t)}\mathrm {d}x\nonumber \\&= {\tilde{Z}}_t^k {\mathrm{e}}^{-k(r+\frac{\gamma ^2}{2})(T-t)+\frac{k^2}{2} \gamma ^2(T-t)}\int _{d_2}^{d_1}\frac{1}{\sqrt{2\pi }}{\mathrm{e}}^{-\frac{1}{2} \underbrace{(x+k\gamma \sqrt{T-t})^2}_{=:y}}\mathrm {d}x\nonumber \\&={\tilde{Z}}_t^k {\mathrm{e}}^{-k(r+\frac{\gamma ^2}{2})(T-t)+\frac{k^2}{2} \gamma ^2(T-t)}\int _{d_2+k\gamma \sqrt{T-t}}^{d_1+k\gamma \sqrt{T-t}} \frac{1}{\sqrt{2\pi }}{\mathrm{e}}^{-\frac{1}{2}y^2}\mathrm {d}y\nonumber \\&= {\tilde{Z}}_t^k {\mathrm{e}}^{-k(r+\frac{\gamma ^2}{2})(T-t)+\frac{k^2}{2}\gamma ^2(T-t)}[N(d_1+k\gamma \sqrt{T-t})-N(d_2+k\gamma \sqrt{T-t})]. \end{aligned}$$

(34)

By application to (13), this leads to

$$\begin{aligned} {\mathbb {E}}[{\tilde{Z}}_TX_T^*]&={\mathbb {E}}\left[ \frac{\left( \frac{ \lambda ^*}{\beta }\right) ^{-1/c_1}{\tilde{Z}}_T^{\frac{c_1-1}{c_1}}}{ \beta }{1}_{\{0<\lambda ^*{\tilde{Z}}_T<v_{\mathrm{M}}'(x^*)\}}\right] \\&\quad -{\mathbb {E}} \left[ \frac{\alpha X_0}{\beta }{\tilde{Z}}_T{1}_{\{0<\lambda ^*{\tilde{Z}}_T< v_{\mathrm{M}}'(x^*)\}}\right] \\&\quad + {\mathbb {E}}\left[ B{\tilde{Z}}_T{1}_{\{0<\lambda ^*{\tilde{Z}}_T<v_{\mathrm{M}}'(x^*)\}} \right] \\&= \left( \frac{\lambda ^*}{\beta }\right) ^{-1/c_1}\frac{1}{\beta }{\mathrm{e}}^{ \frac{1-c_1}{c_1}\left( r+\frac{\gamma ^2}{2}\right) T+\frac{1}{2}\left( \frac{c_1-1}{c_1}\right) ^2\gamma ^2T}\left[ 1-N\left( d+\frac{c_1-1}{c_1} \gamma \sqrt{T}\right) \right] \\&\quad +\left( B-\frac{\alpha X_0}{\beta }\right) {\mathrm{e}}^{-rT}\left[ 1-N(d+\gamma \sqrt{T})\right] =X_0 \end{aligned}$$

as presented in Lemma 1. $\square$

Proof of Proposition 2:

$$\begin{aligned} {\mathbb {E}}\left[ X_T^*\right]&=\frac{1}{\beta }\left( \frac{\lambda ^*}{ \beta }\right) ^{-1/c_1}{\mathbb {E}}\left[ {\tilde{Z}}_T^{-1/c_1}{1}_{\{ \lambda ^*{\tilde{Z}}_T<v_{\mathrm{M}}'(x^*)\}}\right] +\left( B-\frac{\alpha X_0}{\beta }\right) {\mathbb {E}}\left[ {1}_{\{\lambda ^*{\tilde{Z}}_T<v_{\mathrm{M}}'(x^*)\}}\right] \\&\overset{(34)}{=} \frac{1}{\beta }\left( \frac{ \lambda ^*}{\beta }\right) ^{-1/c_1}{\mathrm{e}}^{\frac{1}{c_1}\left( r+\frac{\gamma ^2}{2} \right) T+\frac{1}{2}\frac{\gamma ^2}{c_1^2}T}\left( 1-N\left( d-\frac{\gamma }{c_1} \sqrt{T}\right) \right) \\&\quad +\left( B-\frac{\alpha X_0}{\beta }\right) \left( 1-N\left( d\right) \right) \\ {\mathbb {E}}\left[ \left( X_T^*\right) ^2\right]&= \frac{1}{\beta ^2}\left( \frac{ \lambda ^*}{\beta }\right) ^{-2/c_1}{\mathbb {E}}\left[ {\tilde{Z}}_T^{-2/c_1}{1}_{ \{\lambda ^*{\tilde{Z}}_T<v_{\mathrm{M}}'(x^*)\}}\right] +\left( B-\frac{\alpha X_0}{\beta } \right) ^2 {\mathbb {E}}\left[ {1}_{\{\lambda ^*{\tilde{Z}}_T<v_{\mathrm{M}}'(x^*)\}} \right] \\&\quad + 2\left( B-\frac{\alpha X_0}{\beta }\right) \frac{1}{\beta }\left( \frac{ \lambda ^*}{\beta }\right) ^{-1/c_1}{\mathbb {E}}\left[ {\tilde{Z}}_T^{-1/c_1}{1}_{ \{\lambda ^*{\tilde{Z}}_T<v_{\mathrm{M}}'(x^*)\}}\right] \\&\overset{(34)}{=} \frac{1}{\beta ^2}\left( \frac{ \lambda ^*}{\beta }\right) ^{-2/c_1}{\mathrm{e}}^{\frac{2}{c_1}\left( r+\frac{\gamma ^2}{2} \right) T+2\frac{\gamma ^2}{c_1^2}T}\left( 1-N\left( d-2\frac{\gamma }{c_1}\sqrt{T} \right) \right) \\&\quad + \left( B-\frac{\alpha X_0}{\beta }\right) ^2\left( 1-N\left( d\right) \right) \\&\quad + 2\left( B-\frac{\alpha X_0}{\beta }\right) \frac{1}{\beta }\left( \frac{\lambda ^*}{\beta }\right) ^{-1/c_1}{\mathrm{e}}^{ \frac{1}{c_1}\left( r+\frac{\gamma ^2}{2}\right) T+\frac{1}{2}\frac{\gamma ^2}{c_1^2}T} \left( 1-N\left( d-\frac{\gamma }{c_1}\sqrt{T}\right) \right) \end{aligned}$$

The proposition follows with ${\mathbb {V}}\left[ X_T^*\right] ={\mathbb {E}}\left[ \left( X_T^*\right) ^2\right] - \left( {\mathbb {E}}\left[ X_T^*\right] \right) ^2$. $\square$

Proof of Lemma 2:

Notice that $x^*>B$ and $\lambda ^*{\tilde{Z}}_T<v_{\mathrm{M}}'(x^*)\iff X_T^*>x^*.$ Hence, we can calculate

$$\begin{aligned} \phi _{\mathrm{M}}(\alpha ,\beta )&={\mathbb {E}}\left[ \frac{1}{1-c_1}\left( \alpha X_0+\beta \left( \left( \frac{\left( \frac{\lambda ^*{\tilde{Z}}_T}{ \beta }\right) ^{-1/c_1}-\alpha X_0}{\beta }+B\right) {1}_{\{\lambda ^*{\tilde{Z}}_T<v_{\mathrm{M}}'(x^*)\}}-B\right) ^+ \right) ^{1-c_1}\right] \\&=\frac{1}{1-c_1} {\mathbb {E}}\left[ \left( \alpha X_0+\beta \left( \frac{\left( \frac{\lambda ^*{\tilde{Z}}_T}{\beta }\right) ^{-1/c_1}- \alpha X_0}{\beta }+B-B\right) \right) ^{1-c_1}{1}_{\{\lambda ^*{\tilde{Z}}_T<v_{\mathrm{M}}'(x^*)\}} \right] \\&\quad +\frac{1}{1-c_1} {\mathbb {E}}\left[ \left( \alpha X_0\right) ^{1-c_1}{1}_{ \{\lambda ^*{\tilde{Z}}_T>v_{\mathrm{M}}'(x^*)\}}\right] \\&= \frac{1}{1-c_1}\left( \frac{\lambda ^*}{\beta }\right) ^{\frac{c_1-1}{c_1}}{ \mathbb {E}}\left[ {\tilde{Z}}_T^{\frac{c_1-1}{c_1}}{1}_{\{\lambda ^*{\tilde{Z}}_T<v_{\mathrm{M}}' (x^*)\}}\right] +\frac{1}{1-c_1}(\alpha X_0)^{1-c_1}{\mathbb {E}}\left[ {1}_{\{ \lambda ^*{\tilde{Z}}_T>v_{\mathrm{M}}'(x^*)\}}\right] \\&\overset{(34)}{=} \frac{1}{1-c_1}\left( \frac{ \lambda ^*}{\beta }\right) ^{\frac{c_1-1}{c_1}}{\mathrm{e}}^{\frac{1-c_1}{c_1}\left( r+ \frac{\gamma ^2}{2}\right) T+\left( \frac{c_1-1}{c_1}\right) ^2\frac{\gamma ^2}{2}T} \left( 1-N\left( d+\frac{c_1-1}{c_1}\gamma \sqrt{T}\right) \right) \\&\quad +\frac{1}{1-c_1}(\alpha X_0)^{1-c_1}N(d) \\ \phi _{\mathrm{I}}(\alpha ,\beta )&={\mathbb {E}}\left[ \frac{1}{1-c_2} \left( X_T^*-\beta \left( X_T^*-B\right) ^+\right) ^{1-c_2}\right] \\&={\mathbb {E}}\left[ \frac{1}{1-c_2}\left( \left( \left( \frac{\left( \frac{ \lambda ^*{\tilde{Z}}_T}{\beta }\right) ^{-1/c_1}-\alpha X_0}{\beta }+B \right) \left( 1-\beta \right) +\beta B\right) {1}_{\{\lambda ^*{\tilde{Z}}_T< v_{\mathrm{M}}'(x^*)\}}\right) ^{1-c_2}\right] \\&=\frac{1}{1-c_2} {\mathbb {E}}\left[ \left( (1-\beta ) \frac{\left( \frac{\lambda ^*{\tilde{Z}}_T}{\beta }\right) ^{-1/c_1}-\alpha X_0}{\beta }+B\right) ^{1-c_2}{1}_{\{\lambda ^*{\tilde{Z}}_T<v_{\mathrm{M}}'(x^*)\}}\right] \end{aligned}$$

$\square$

Proof of Lemma 3:

Let $\alpha , \beta$ and the optimal payoff $X_T(\alpha ,\beta )$ be given. Now, let us choose $(\alpha ',\beta ')>(\alpha ,\beta )$, but the same $X_T(\alpha ,\beta )$. Then, the budget constraint is fulfilled and

$$\begin{aligned} U_{\mathrm{M}}(M_T(\alpha ',\beta ',X_T(\alpha ,\beta )))>U_{\mathrm{M}}(M_T(\alpha , \beta ,X_T(\alpha ,\beta )))=u_{\mathrm{M}}(X_T(\alpha ,\beta )). \end{aligned}$$

After taking expectations and then optimizing the left-hand side in $X_T$ to find the optimal $X_T(\alpha ',\beta ')$, we learn that the left-hand side must increase, giving us $\phi _{\mathrm{M}}(\alpha ',\beta ')>\phi _{\mathrm{M}}(\alpha ,\beta )$. $\square$

Proof of Proposition 3

Let $\alpha >0$ be arbitrary, but fixed. The main idea of the proof is to show that for a small $\beta _2$ and $\beta _1\rightarrow 0$ the optimal payoff $X_{\beta _1}^*$ is first-order stochastically dominated by $X_{\beta _2}^*$ which means that ${\mathbb {E}}[u(X_{\beta _1})]\le {\mathbb {E}}[u(X_{\beta _2})]$ for all increasing utility functions u. The proof is structured in six parts. We show:

• (1) $x_{\beta }^*$ is a decreasing and $v_{\mathrm{M}}'(x_{\beta }^*)$ is an increasing function in ${\beta }$.

• (2) $\lambda _{\beta }^*$ is an increasing function in $\beta$.

• (3) $\frac{v_{\mathrm{M}}'(x_{\beta }^*)}{\beta }$ converges to a constant for $\beta \rightarrow 0$.

• (4) $\frac{\lambda _{\beta }^*}{\beta }$ and $\frac{\lambda _{\beta }^*}{v_{\mathrm{M}}'(x_{\beta }^*)}$ are decreasing functions in $\beta$, for small $\beta$.

• (5) Estimates for help functions.

• (6) Stochastic dominance of $X_{\beta _2}^*$ over $X_{\beta _1}^*$, for small $\beta _2$ and $\beta _1\rightarrow 0$.

(1) $x_{\beta }^*$ is a decreasing and $\mathbf {v_{{M}}'}{(\mathbf {x}_{\beta }^*)}$ is an increasing function in ${\beta }$.

Let us define $u_{\mathrm{M}}^{\beta }(x)=\frac{1}{1-c_1}\left( \alpha X_0+\beta (x-B)^+\right) ^{1-c_1}$, $x>0$ arbitrary. It follows:

• (i) $u_{\mathrm{M}}^{\beta _1}(x)<u_{\mathrm{M}}^{\beta _2}(x) \quad \forall \beta _1<\beta _2, \forall x>B$

• (ii) $\frac{\partial }{\partial x}u_{\mathrm{M}}^{\beta _1}(x)>\frac{\partial }{\partial x}u_{\mathrm{M}}^{\beta _2}(x) \quad \forall \beta _1<\beta _2, \forall x>B$

• (iii) $\frac{\partial }{\partial x}u_{\mathrm{M}}^{\beta }> 0\quad \forall x>B$

• (iv) $\frac{\partial ^2}{\partial ^2 x}u_{\mathrm{M}}^{\beta }(x)< 0\quad \forall x>B$

Let $\beta _1<\beta _2$. From the definition of $x^*$, we know:

$$\begin{aligned}&u_{\mathrm{M}}^{\beta _1}(0)+\frac{\partial }{\partial x}u_{\mathrm{M}}^{\beta _1}(x)|_{x_{\beta _1}^*}\cdot x_{\beta _1}^*=u_{\mathrm{M}}^{\beta _1}(x_{\beta _1}^*)\\&u_{\mathrm{M}}^{\beta _2}(0)+\frac{\partial }{\partial x}u_{\mathrm{M}}^{\beta _2}(x)|_{x_{\beta _2}^*}\cdot x_{\beta _2}^*=u_{\mathrm{M}}^{\beta _2}(x_{\beta _2}^*) \end{aligned}$$

First, we show that $x_{\beta }^*$ is decreasing in $\beta$. We assume $x_{\beta _1}^*<x_{\beta _2}^*$. Then, it is

$$\begin{aligned} u_{\mathrm{M}}^{\beta _2}(0)+\frac{\partial }{\partial x}u_{\mathrm{M}}^{\beta _2}(x)|_{x_{\beta _2}^*}\cdot x_{\beta _1}^*>u_{\mathrm{M}}^{\beta _2}(x_{\beta _1}^*)\overset{\text {(i)}}{>}u_{\mathrm{M}}^{\beta _1}(x_{\beta _1}^*)=u_{\mathrm{M}}^{\beta _1}(0)+\frac{\partial }{\partial x}u_{\mathrm{M}}^{\beta _1}(x)|_{x_{\beta _1}^*}\cdot x_{\beta _1}^*\\ \overset{u_{\mathrm{M}}^{\beta _1}(0)=u_{\mathrm{M}}^{\beta _2}(0)}{\iff }\frac{\partial }{\partial x} u_{\mathrm{M}}^{\beta _2}(x_{\beta _2}^*)>\frac{\partial }{\partial x} u_{\mathrm{M}}^{\beta _1}(x_{\beta _1}^*)\overset{\text {(iv)}}{>}\frac{\partial }{\partial x}u_{\mathrm{M}}^{\beta _1}(x_{\beta _2}^*)\overset{\text {(ii)}}{>}\frac{\partial }{\partial x} u_{\mathrm{M}}^{\beta _2}(x_{\beta _2}^*) \end{aligned}$$

which is a contradiction. Hence, it must be $x_{\beta _1}^*>x_{\beta _2}^*$, i.e., $x_{\beta }^*$ is decreasing in $\beta$. Let $v_{\mathrm{M}}$ be the concavification of $u_{\mathrm{M}}^{\beta }$. Since the concavification $v_{\mathrm{M}}$ is a concave function, it follows $v_{\mathrm{M}}'(x_{\beta _2}^*)>v_{\mathrm{M}}'(x_{\beta _1}^*)$, such that $v_{\mathrm{M}}'(x_{\beta }^*)$ is an increasing function of $\beta$.

(2) $\lambda _{\beta }^*$ is an increasing function in $\beta$.

We look at Eq. (15):

$$\begin{aligned}&\overbrace{\underbrace{\left( \frac{\lambda _{\beta }^*}{\beta }\right) ^{-1/c_1} \frac{1}{\beta }}_{\text{ Decreasing } \text{ in } \lambda _{\beta }^*}}^{\text{ Increasing } \text{ in } \beta \text{(for } c_1<1\text{) }}\cdot E_1\cdot \overbrace{\underbrace{\left[ 1-N\left( d\left( \frac{\lambda _{\beta }^*}{ v_{\mathrm{M}}'(x_{\beta }^*)}\right) +\varepsilon _1\right) \right] }_{\text{ Decreasing } \text{ in } \lambda _{\beta }^*}}^{\text{ Increasing } \text{ in } \beta }\nonumber \\&+\underbrace{\left( B-\frac{\alpha X_0}{\beta }\right) }_{\text{ Increasing } \text{ in } \beta }\cdot E_2\cdot \underbrace{\left[ 1-N\left( d\left( \frac{\lambda _{\beta }^*}{v_{\mathrm{M}}'(x_{ \beta }^*)}\right) +\varepsilon _2\right) \right] }_{\text{ Increasing } \text{ in } \beta \text{ and } \text{ decreasing } \text{ in } \lambda _{\beta }^*}=\underbrace{X_0}_{\text{ Constant }} \end{aligned}$$

(35)

The right-hand side of the upper equation is constant while the left-hand side appears to be increasing in $\beta$ (remember that $v_{\mathrm{M}}'(x_{\beta }^*)$ is increasing in $\beta$) for $c_1<1$ (while assuming $\lambda _{\beta }^*$ to be constant) and simultaneously decreasing in $\lambda _{\beta }^*$ (while assuming $\beta$ to be constant). Therefore, $\lambda _{\beta }^*$ is increasing in $\beta$. This implies that $\frac{\lambda _{\beta }^*}{\beta }$ is monotone for small $\beta$ which is relevant for Step 4).

(3) ${\frac{{\mathbf {v_M'}}(x_{\beta }^*)}{\beta }}$ converges to a constant for $\beta {\rightarrow }$ 0.

Here we show that $v_{\mathrm{M}}'(x_{\beta }^*)\sim \beta$, which means $\lim \nolimits _{\beta \rightarrow 0}\frac{v_{\mathrm{M}}'(x_{\beta }^*)}{\beta }=a$ where $a>0$ is a constant.

From (12), the definition of $x_{\beta }^*$, we know

$$\begin{aligned}&\frac{1}{1-c_1}\left[ (\alpha X_0+\beta (x_{\beta }^*-B))^{1-c_1}-(\alpha X_0)^{1-c_1}\right] =\beta x_{\beta }^*\cdot (\alpha X_0+\beta (x_{\beta }^*-B))^{-c_1}\\\iff & {} \left[ \frac{\alpha X_0+(\beta x_{\beta }^*-\beta B)-(\alpha X_0+(\beta x_{\beta }^*-\beta B))^{c_1}\cdot (\alpha X_0)^{1-c_1}}{\beta x_{\beta }^*}\right] =1-c_1. \end{aligned}$$

For $\beta \rightarrow 0$, there are three different possible cases for $\beta x_{\beta }^*$.

• $\beta x_{\beta }*\rightarrow \infty$

• $\beta x_{\beta }*\rightarrow 0$

• $\beta x_{\beta }*\rightarrow {\hbox {const.}}>0$

For the first two cases, we differentiate the left side of the upper equation w.r.t. $(\beta x_{\beta }*)$ and apply the rule of l’Hopital. Hence, for $\beta \rightarrow 0$ it must hold

$$\begin{aligned} \lim \limits _{\beta \rightarrow 0}\frac{1-(\alpha X_0)^{1-c_1}\cdot c_1\cdot (\alpha X_0+(\beta x_{\beta }^*-\beta B))^{c_1-1}}{1}=1-c_1. \end{aligned}$$

• For $\beta x_{\beta }*\rightarrow \infty$, this leads to $1\ne 1-c_1$.

• For $\beta x_{\beta }*\rightarrow 0,$ this leads to $1-c_1=1-c_1$. $\checkmark$

The third case, $\beta x_{\beta }*\rightarrow {\hbox {const.}}>0$, cannot be true, because the unique limit cannot be 0 and a constant greater than 0 simultaneously. Therefore, $\lim \nolimits _{\beta \rightarrow 0}\beta x_{\beta }^*= 0$.

Since $v_{\mathrm{M}}'(x_{\beta }^*)=\beta \cdot (\alpha X_0+\beta (x_{\beta }^*-B))^{-c_1}$, we obtain

$$\begin{aligned} \lim \limits _{\beta \rightarrow 0}\frac{v_{\mathrm{M}}'(x_{\beta }^*)}{\beta }=\lim \limits _{\beta \rightarrow 0}(\alpha X_0+\beta x_{\beta }^*-\beta B)^{-c_1}=(\alpha X_0)^{-c_1} \end{aligned}$$

where $(\alpha X_0)^{-c_1}$ is constant. Hence, $v_{\mathrm{M}}'(x_{\beta }^*)\sim \beta$ as prescribed.

(4) ${\frac{{\varvec{\lambda }}_{\beta }^*}{\beta }}$ and ${\frac{{\varvec{\lambda }}_{\beta }^*}{{\mathbf {v_M}}'({\mathbf {x}}_{\beta }^*)}}$ are decreasing functions in ${\beta }$, for small ${\beta }$.

We define $y_{\beta }=\frac{\lambda _{\beta }^*}{\beta }$ and use the result from Step 3)

$$\begin{aligned} \lim \limits _{\beta \rightarrow 0}\left( \frac{\lambda _{\beta }^*}{v_{\mathrm{M}}'(x_{\beta }^*)}\right) =a\cdot \lim \limits _{\beta \rightarrow 0}\left( \frac{\lambda _{\beta }^*}{\beta }\right) \text{ with } a>0. \end{aligned}$$

(36)

Furthermore, let

$$\begin{aligned} E_3={\mathrm{e}}^{-\frac{c_1-1}{c_1}\left( r+\frac{\gamma ^2}{2} \right) T+\frac{1}{2}\left( \frac{c_1-1}{c_1}\right) ^2\gamma ^2T}, \varepsilon _1=\frac{c_1-1}{c_1}\gamma \sqrt{T}, E_4={\mathrm{e}}^{-rT}, \varepsilon _2=\gamma \sqrt{T} \end{aligned}$$

and

$$\begin{aligned} d\left( \frac{\lambda ^*}{v_{\mathrm{M}}'(x_{\beta }^*)}\right) =\frac{{\hbox {log}}\left( \frac{ \lambda _{\beta }^*}{v_{\mathrm{M}}'(x_{\beta }^*)}\right) -(r+\frac{\gamma ^2}{2})T}{\gamma \sqrt{T}}. \end{aligned}$$

Then, using (15) we get

$$\begin{aligned} L(\beta ):= & {} E_3\cdot y_{\beta }^{-1/c_1}\cdot \left[ 1-N\left( d\left( \frac{\lambda _{\beta }^*}{v_{\mathrm{M}}'(x_{ \beta }^*)}\right) +\varepsilon _1\right) \right] -\alpha X_0\cdot E_4\cdot \left[ 1-N\left( d\left( \frac{\lambda _{\beta }^*}{v_{\mathrm{M}}'(x_{\beta }^*)}\right) + \varepsilon _2\right) \right] \nonumber \\= & {} \beta \cdot \left( X_0-E_4\cdot B\cdot \left[ 1-N\left( d\left( \frac{\lambda _{\beta }^*}{v_{\mathrm{M}}'(x_{\beta }^*)} \right) +\varepsilon _2\right) \right] \right) =:R(\beta ).\nonumber \\ \overset{(36)}{\Rightarrow }\lim \limits _{\beta \rightarrow 0}L(\beta )= & {} E_3\cdot \lim \limits _{\beta \rightarrow 0}\left( \frac{1-N(d(a\cdot y_{\beta })+\varepsilon _1)}{y_{\beta }^{1/c_1}}\right) -\alpha X_0 \cdot E_4\cdot \lim \limits _{\beta \rightarrow 0}\left( 1-N(d(a\cdot y_{\beta })+\varepsilon _2)\right) \nonumber \\= & {} \lim \limits _{\beta \rightarrow 0}\left( \beta \cdot (X_0-E_4\cdot B\cdot [1-N(d(a\cdot y_{\beta })+\varepsilon _2)])\right) =\lim \limits _{\beta \rightarrow 0}R(\beta ) \end{aligned}$$

(37)

Looking at Eq. (37) and recalling that $y_{\beta }$ is monotone for small $\beta$, we obtain:

• Assume $y_{\beta }\downarrow 0$: Then, $\lim \nolimits _{\beta \rightarrow 0}L(\beta )=E_3\cdot \infty -\alpha X_0\cdot E_4\ne 0=\lim \nolimits _{\beta \rightarrow 0}R(\beta )\Rightarrow$ Contradiction to (37)

• Assume $y_{\beta }\uparrow \infty$: Then, $\lim \nolimits _{\beta \rightarrow 0}L(\beta )=E_3\cdot 0-0=0=\lim \nolimits _{\beta \rightarrow 0}R(\beta )$ $\checkmark$

Analogously to the argumentation in Step 3, $y_{\beta }\rightarrow y^*>0$ cannot hold, because $y_{\beta }$ cannot converge to $\infty$ and a constant greater than 0 simultaneously. Hence, $\lim \nolimits _{\beta \rightarrow 0}y_{\beta }=\infty$, and thus, for small $\beta$, $\frac{\lambda _{\beta }^*}{\beta }$ and $\frac{\lambda _{\beta }^*}{v_{\mathrm{M}}'(x_{\beta }^*)}$ are decreasing in $\beta$.

(5) Estimates for help functions.

Let $f(\beta ):=\left( \frac{\lambda _{\beta }^*}{\beta }\right) ^{-1/c_1}\cdot \frac{1}{\beta }$. Using Eq. (15) and the fact that $\frac{\lambda _{\beta }^*}{v_{\mathrm{M}}'(x_{\beta }^*)}$ is decreasing in $\beta$ for small $\beta$, the terms $\left[ 1-N\left( d\left( \frac{\lambda _{\beta }^*}{v_{\mathrm{M}}'(x_{\beta }^*)}\right) +\varepsilon _1\right) \right]$, $\left[ 1-N\left( d\left( \frac{\lambda _{\beta }^*}{v_{\mathrm{M}}'(x_{\beta }^*)}\right) +\varepsilon _2\right) \right]$ and $\left( B-\frac{\alpha X_0}{\beta }\right)$ are increasing in $\beta$. It follows that $f(\beta )$ is decreasing in $\beta$ for small $\beta$.

Let $\beta _1<\beta _2$ be small and define $g(\beta _1,\beta _2):=\left( \frac{\lambda _{\beta _2}^*}{\lambda _{\beta _1}^*}\cdot \frac{\beta _1}{\beta _2}\right) ^{1/c_1}$. Then, it holds

• (I) $g(\beta _1,\beta _2)<1$, as direct result of $\frac{\lambda _{\beta _2}^*}{\beta _2}<\frac{\lambda _{\beta _1}^*}{\beta _1}$.

• (II) $\beta _1-\beta _2\cdot g(\beta _1,\beta _2)<0\iff \frac{\beta _1}{\beta _2}<g(\beta _1,\beta _2)$ which follows from $f(\beta _2)<f(\beta _1)$.

(6) Stochastic dominance of $X_{\beta _2}^*$ over $X_{\beta _1}^*$, for small $\beta _2$ and $\beta _1\rightarrow 0$.

In the final step, we now look at the cumulative distribution function of the optimal payoff

$$\begin{aligned} X_{\beta }^*=\left[ \frac{\left( \frac{\lambda _{\beta }^*{\tilde{Z}}_T}{\beta }\right) ^{-1/c_1}-\alpha X_0}{\beta }+B\right] {1}_{\{\lambda _{\beta }^*{\tilde{Z}}_T<v_{\mathrm{M}}'(x_{\beta }^*)\}}. \end{aligned}$$

Remember that $X_{\beta }^*$ never takes values between 0 and $x_{\beta }^*>B$. Let $x>x_{\beta }^*$ be arbitrary. Then:

$$\begin{aligned} X_{\beta }^*\le x&\iff \left( \frac{\beta }{\lambda _{\beta }^*{\tilde{Z}}_T}\right) ^{1/c_1}\le \beta (x-B)+\alpha X_0 \\&\iff \frac{\beta }{\lambda _{\beta }^*{\tilde{Z}}_T}\le (\alpha X_0+\beta (x-B))^{c_1}\\&\iff {\tilde{Z}}_T\ge \frac{\beta }{\lambda _{\beta }^*}\cdot (\alpha X_0+\beta (x-B))^{-c_1}=:h_{\beta }(x) \end{aligned}$$

Hence, for any $x>x_{\beta }^*$ the cumulative distribution function is given by

$$\begin{aligned} F_{X_{\beta }^*}(x)={\mathbb {P}}\left( {\tilde{Z}}_T\ge h_{\beta }(x)\right) . \end{aligned}$$

Let $\beta _1<\beta _2$ be small and $x\ge x_{\beta _1}^*>x_{\beta _2}^*$ (where the last inequality comes from Step 1). Then,

$$\begin{aligned} F_{X_{\beta _1}^*}(x)\ge F_{X_{\beta _2}^*}(x)&\iff h_{\beta _1}(x)\le h_{\beta _2}(x)\\&\iff \frac{\beta _1}{\lambda _{\beta _1}^*}\cdot (\alpha X_0+\beta _1(x-B))^{-c_1}\le \frac{\beta _2}{\lambda _{\beta _2}^*}\cdot (\alpha X_0+\beta _2(x-B))^{-c_1}\\&\iff \frac{\lambda _{\beta _2}^*}{\beta _2}\cdot (\alpha X_0+\beta _2(x-B))^{c_1}\le \frac{\lambda _{\beta _1}^*}{\beta _1}\cdot (\alpha X_0+\beta _1(x-B))^{c_1}\\&\iff \left( \frac{\lambda _{\beta _2}^*}{\beta _2}\right) ^{1/c_1}\cdot (\alpha X_0+\beta _2(x-B))\le \left( \frac{\lambda _{\beta _1}^*}{\beta _1}\right) ^{1/c_1}\cdot (\alpha X_0+\beta _1(x-B))\\&\iff g(\beta _1,\beta _2)\cdot (\alpha X_0+\beta _2(x-B))\le \alpha X_0+\beta _1(x-B)\\&\iff (g(\beta _1,\beta _2)-1)\cdot \alpha X_0\le (\beta _1-\beta _2g(\beta _1,\beta _2))(x-B)\\&\iff x\le B+\frac{g(\beta _1,\beta _2)-1}{\beta _1-\beta _2\cdot g(\beta _1,\beta _2)}\cdot \alpha X_0=:S(\beta _1,\beta _2) \end{aligned}$$

Now we show that $S(\beta _1,\beta _2)\rightarrow \infty$ for $\beta _1\rightarrow 0$.

$$\begin{aligned} \frac{1-g(\beta _1,\beta _2)}{\beta _2\cdot g(\beta _1,\beta _2)-\beta _1} \uparrow \infty \text{ for } \beta _1\rightarrow 0&\iff \frac{\beta _2\cdot g(\beta _1,\beta _2)-\beta _1}{1-g(\beta _1,\beta _2)}\downarrow 0 \text{ for } \beta _1\rightarrow 0 \nonumber \\&\iff -\beta _2+\frac{\beta _2-\beta _1}{1-g(\beta _1,\beta _2)}\downarrow 0 \text{ for } \beta _1\rightarrow 0 \end{aligned}$$

(38)

In Step 4, we have seen that $\frac{\lambda _{\beta }^*}{\beta }\rightarrow \infty$ for $\beta \rightarrow 0$. Hence, $\frac{\beta _1}{\lambda _{\beta _1}^*}\downarrow 0$ for $\beta _1\rightarrow 0$. Therefore, we know

$$\begin{aligned} g(\beta _1,\beta _2)\downarrow 0 \text{ for } \beta _1\rightarrow 0. \end{aligned}$$

(I) and (II) in Step 5 imply that $-\beta _2+\frac{\beta _2-\beta _1}{1-g(\beta _1,\beta _2)}$ is always positive. Thus, Eq. (38) converges to 0 from above for $\beta _1\rightarrow 0$, such that $S(\beta _1,\beta _2) \rightarrow \infty$ for $\beta _1\rightarrow 0$ and small $\beta _2$.

Hence, for $\beta _1\rightarrow 0$ and small $\beta _2$ it holds

$$\begin{aligned} F_{X_{\beta _1}^*}(x)\ge F_{X_{\beta _2}^*}(x) \quad \forall x\in \varOmega _1=[x_{\beta _1}^*,\infty ), \end{aligned}$$

i.e., $X_{\beta _2}^*$ dominates $X_{\beta _1}^*$ on $\varOmega _1$ by first-order stochastic dominance. Therefore,

$$\begin{aligned} {\mathbb {E}}\left[ {1}_{\varOmega _1}(X_{\beta _1}^*)\cdot u(X_{\beta _1}^*)\right] \le {\mathbb {E}}\left[ {1}_{\varOmega _1}(X_{\beta _2}^*)\cdot u(X_{\beta _2}^*)\right] \quad \forall \text{ increasing } \text{ functions } u. \end{aligned}$$

We thus conclude

$$\begin{aligned} {\mathbb {E}}\left[ u(X_{\beta _1}^*)\right] ={\mathbb {E}}\left[ {1}_{ \varOmega _1}(X_{\beta _1}^*)\cdot u(X_{\beta _1}^*)\right] \le {\mathbb {E}}\left[ {1}_{\varOmega _1}(X_{\beta _2}^*)\cdot u(X_{\beta _2}^*)\right] \le {\mathbb {E}}\left[ u(X_{\beta _2}^*)\right] . \end{aligned}$$

$\square$

Proof of Lemma 4:

Let $({\hat{\alpha }},{\hat{\beta }})\notin P$. Set $\phi _\mathrm{min}= \phi _{\mathrm{M}}({\hat{\alpha }},{\hat{\beta }})$ and let $(\alpha ,\beta )$ be the solution of the corresponding problem (28). As $({\hat{\alpha }},{\hat{\beta }})$ is admissible for (28) and $(\alpha ,\beta )$ the optimal solution, we conclude

$$\begin{aligned} \phi _{\mathrm{I}}(\alpha ,\beta )\ge & {} \phi _{\mathrm{I}}({\hat{\alpha }},{\hat{\beta }}). \end{aligned}$$

As $({\hat{\alpha }},{\hat{\beta }})\notin P$, we know that $({\hat{\alpha }},{\hat{\beta }})$ cannot be an optimal solution for (28). Hence, the inequality must be strict. $\square$